Question

for the dy/dx of xy^3+xy=16

is the dy/dx= y^3/3xy^2+x??

Answer 1

hmm

x'.y^3 + x.3y^2.y' + x'.y + x.y' = 0

x.3y^2.y' + x.y' = -y^3 -y

y'(3xy^2 +x) = -y^3 -y

y' = \(\frac{-y^3 -y}{3xy^2 +x}\)